Pi, momentum, and kinetic energy

There is a geometric way one can show the relation between kinetic energy and momentum which is a mathematical curiosity in my opinion. Let the mass of an object be equal to 2 PI. Then:

P = 2 pi v

KE = 1/2 (2 pi ) v sq
or
KE = pi v sq

Hence, graphically, if we set v = r, where r is the radius of a circle, we have a circle whose circumference is momentum, 2 pi v, and whose area is the kinetic energy, pi v sq. Thus, for the case where the mass is 2 pi, we see the kinetic energy is geometrically bounded by the momentum of the object. Interesting.
Wonder what would happen to this circle at relativistic velocities?

There is a geometric way one can show the relation between kinetic energy and momentum which is a mathematical curiosity in my opinion. Let the mass of an object be equal to 2 PI. Then:

P = 2 pi v

KE = 1/2 (2 pi ) v sq
or
KE = pi v sq

Hence, graphically, if we set v = r, where r is the radius of a circle, we have a circle whose circumference is momentum, 2 pi v, and whose area is the kinetic energy, pi v sq. Thus, for the case where the mass is 2 pi, we see the kinetic energy is geometrically bounded by the momentum of the object. Interesting.
Wonder what would happen to this circle at relativistic velocities?

I personally don't think there is any significance here. For the cases where v is not equal to r, the properties of a circle disappear. There is no fundamental property here that can be generalized to all cases of v, and is nothing more than a result of simply choosing the variables to get the desired outcome.

I personally don't think there is any significance here. For the cases where v is not equal to r, the properties of a circle disappear. There is no fundamental property here that can be generalized to all cases of v, and is nothing more than a result of simply choosing the variables to get the desired outcome.

I'm a little agree with it, for the special case won't exist if the special amount ##\pi## is replaced with any others.